Let $V$ be a normed vector space and $K$ a compact subset of $V$. Is the convex hull of $K$ given by $$\langle K \rangle = \left\{\sum_{i=1}^n t_i x_i\mid x_i\in K, t_i≥0\text{ s.t. }\sum_i t_i=1\right\}$$ again compact?

  • $\begingroup$ Possibly related: mathoverflow.net/questions/156321/convex-hulls-of-compact-sets. $\endgroup$ – Martin R Nov 16 '16 at 17:47
  • $\begingroup$ @MartinR thanks, from the link in that question, the answer is no. $\endgroup$ – s.harp Nov 16 '16 at 17:54
  • $\begingroup$ I don't think your definition of convex hull is correct. If $K=\{x,y,z\}$, is $\frac{x+y+z}{3} \in \langle K \rangle$? $\endgroup$ – Thomas Aug 25 '17 at 22:38
  • $\begingroup$ @Thomas you are right. We obviously want want $\sum_i t_i\,x_i$ where $\sum_i t_i =1, t_i≥0$ or with what I have written for $\langle\cdot\rangle$: $$\bigcup_n \underbrace{\langle ... \langle}_{n} K\underbrace{\rangle...\rangle}_{n}$$Alternatively the smallest set containing $K$ so that $\langle X\rangle =X$ with the definition of $\langle \cdot\rangle $ that I have written. $\endgroup$ – s.harp Aug 27 '17 at 12:09

Answering the question with the counter example from the link in the math overflow question linked by Martin R:

Consider $$u_n=(\underbrace{0,...,0}_{n-1},1/n,0,...)$$ and $K=\bigcup_n \{u_n\} \cup \{0\}$ a compact subset of $\mathscr l^p(\mathbb N)$. The convex hull of $K$ is given by elements of the form: $$\sum_{n=1}^k a_n u_{n}\qquad\text{s.t.:}\quad \sum_{n=1}^k a_n≤1\qquad a_n≥0$$ So also $\sum_{n=1}^k 2^{-n}u_n$ lies in it. But this sequence converges to $\sum_{n=1}^\infty 2^{-n}u_n $ which does not lie in it.

However: From Theorem 5.35: The closed convex hull is compact in a complete normed vector space. So the convex hull of a compact set is pre-compact (or totally bounded if the original space is not complete).


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